1. 12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

2. Determining the Structure of an Organic Compound
The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants
In the 19th and early 20th centuries, structures were determined by synthesis and chemical degradation that related compounds to each other
Physical methods now permit structures to be determined directly. We will examine:
mass spectrometry (MS)
infrared (IR) spectroscopy
nuclear magnetic resonance spectroscopy (NMR)
ultraviolet-visible spectroscopy (VIS)

3. Why this Chapter?
Finding structures of new molecules synthesized is critical
To get a good idea of the range of structural techniques available and how they should be used

4. 12.1 Mass Spectrometry of Small Molecules:Magnetic-Sector Instruments
Bonds in cation radicals begin to break (fragment)
Sample vaporized and subjected to bombardment by electrons that remove an electron creating a cation radical
Mass to charge (m/z) ratio is measured
Molecular ion (M+)
shows molecular weight

5. The Mass Spectrum
Plot mass of ions (m/z) (x-axis) versus the intensity of the signal (roughly corresponding to the number of ions) (y-axis)
Tallest peak is base peak (100%)
Other peaks listed as the % of that peak
Peak that corresponds to the unfragmented radical cation is parent peak or molecular ion (M+)
Propane
MW = M+ = 44

6. 12.2 Interpreting Mass Spectra
Molecular weight from the mass of the molecular ion
Double-focusing instruments provide high-resolution “exact mass”
atomic mass units – distinguishing specific atoms
Example MW “72” is ambiguous: C5H12 and C4H8O but:
C5H amu exact mass C4H8O amu exact mass
Result from fractional mass differences of atoms 16O = , 12C = , 1H =
Instruments include computation of formulas for each peak
If parent ion not present due to electron bombardment causing breakdown, “softer” methods such as chemical ionization are used

7. M+ peak Example:
Possible formulas can be determined from the mass. (M+)
Predict possible molecular formulas containing C, H, and maybe O if :
M+ = 86
M+ = 156

8. M+ peak Example: C6H14 C5H10O C4H6O2 C3H2O3 C11H24 C12H12 C11H8O
Possible formulas can be predicted from the mass. (M+)
Predict possible molecular formulas containing C, H, and maybe O if :
M+ = 86
M+ = 156
C6H14
C5H10O C4H6O2
C3H2O3
C11H24
C12H12 C11H8O
C10H20O
C10H4O2 C9H16O2
C8H12O3
C7H8O4
C6H4O5

9. M+ peak :
Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample
(M+1) from 1.1% 13C in nature
(M and M+2) in 75.8%/24.2% ratio = 35Cl and 37Cl
(M and M+2) in 50.7%/49.3% ratio = 79Br and 81Br
Propane
MW = M+ = 44
M+1 = 45
From 1.1% 13C

10. M+ peak: Halides M+ and M+2 in 75.8%:24.2% (~ 3:1) ratio
Click on image to enlarge
M+ peak: Halides
M+ and M+2 in 75.8%:24.2% (~ 3:1) ratio
= 35Cl and 37Cl
M+ and M+2 in 50.7%:49.3% (~ 1:1) ratio
= 79Br and 81Br
CH3Cl

11. Mass-Spec Fragmentation Patterns
Molecular ions break into characteristic fragments that can be identifed
Serves as a “fingerprint” for comparison with known materials in analysis (used in forensics)
Positive charge goes to fragments that best can stabilize it

12. Mass-Spec Fragmentation Patterns
CH3=15
MW=72;
M+ peak not seen

13. Mass Spec: Fragmentation of Hexane
Hexane (m/z = 86 for parent) has peaks at
m/z = 71, 57, 43, 29
71=Loss of CH3 (15)
57=Loss of CH3CH2 (29)

14. Learning Check:
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

15. Solution:
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.
98-69=29 loss of (CH3CH2)
98-83=15 (loss of CH3)
M+ =98

16. Learning Check:
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

17. Solution:
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.
98-83=15 (loss of CH3)
M+ =98

18. 12.3 Mass Spec: Common Functional Groups
Alcohols:
Alcohols undergo -cleavage (at bond next to the C-OH) as well as loss of H-OH (18) to give C=C
Loss of 18

19. Mass Spec: Alcohols

20. Mass Spec: Amines Nitrogen Rule: Amines with odd # of N’s have Odd M+
Amines undergo -cleavage, generating radicals

21. Mass Spec: Carbonyl Compounds
A C-H that is three atoms away leads to an internal transfer of a proton to the C=O, called the McLafferty rearrangement
Loss of 28
Carbonyl compounds can also undergo  cleavage

22. 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments
Most biochemical analyses by MS use:
electrospray ionization (ESI)
Matrix-assisted laser desorption ionization (MALDI)
• Linked to a time-of-flight mass analyzer
MALDI-TOF MS of chicken eg-white lysozyme

23. 12.5 Spectroscopy and the Electromagnetic Spectrum
Radiant energy is proportional to its frequency (cycles/s = Hz) as a wave (Amplitude is its height)
Different types are classified by frequency or wavelength ranges

24. Electromagnetic Spectrum
High frequency (u) Short wavelength (l)
= High energy
Low frequency (u)
Long wavelength (l)
= Low energy

25. Absorption Spectra
Organic compound exposed to electromagnetic radiation, can absorb energy of certain wavelengths.
Changing wavelengths to determine which are absorbed and which are transmitted produces an absorption spectrum
Energy absorbed is shown as dips in spectrum
High frequency (u) Short wavelength (l)
= High energy
Infrared Absorption of Ethyl Alcohol CH3CH2OH
Low frequency (u)
Long wavelength (l)
= Low energy

26. 12.6 Infrared Spectroscopy
High frequency = High E
Low frequency = Low E
IR region lower energy than visible light (< red – produces heating as with a heat lamp)
2.5  106 m to 2.5  105 m region used by organic chemists for structural analysis
IR energy in a spectrum is usually measured as wavenumber (cm-1), the inverse of wavelength and proportional to frequency
Specific IR absorbed by organic molecule related to its structure

27. Infrared Spectroscopy
IR energy absorption corresponds to atomic movements, such as vibrations and rotations from bending and stretching of bonds between groups of atoms
Energy is characteristic of the bonding of atoms in a functional group
Bond stretching dominates higher energy modes
Light objects connected to heavy objects vibrate fastest: C-H, N-H, O-H
For two heavy atoms, stronger bond requires more energy: C º C, C º N > C=C, C=O, C=N > C-C, C-O, C-N, C-halogen

28. 12.7 Interpreting Infrared Spectra
Most functional groups absorb at a characteristic energy
cm-1 N-H, C-H, O-H (stretching)
N-H, O-H
3000 C-H
cm-1 double bonds C=C, C=O, C=N (stretching)
C=O
C=C cm-1
Below 1500 cm-1 “fingerprint” Single bonds C-C, C-O, C-N, C-X (vibrations)
cm-1 CºC and C º N (stretching)

29. Infrared Spectra: Functional Grps
Characteristic higher energy IR absorptions used to confirm the existence of the presence of functional groups

30. 12.8 IR Spectra: Functional Grps
Alkane
C-C
-C-H
Alkene
Alkyne

31. IR: Aromatic Compounds
(Subsituted benzene “teeth”)
C≡C

32. IR: Alcohols and Amines
O-H broadens with Hydrogen bonding
CH3CH2OH
C-O
N-H broadens with Hydrogen bonding
Amines similar to OH

33. IR: Alcohols: O-H stretch
Gas phase
(no H-bonding)
CCl4 sln (0.25M)
(some H-bonding)
Liquid Film
(Lots of H-bonding)

34. IR: Alcohols
C-O

35. IR: Amines Amines similar to OH N-H broadens with Hydrogen bonding
Primary and secondary amines exhibit a characteristic broad IR N-H stretching absorption between 3250 and 3500 cm-1.
Primary amines show two strong peaks in this range, whereas secondary amines show only one.
Primary amines also show a band near 1600 cm-1 due to a scissoring motion of the NH2 group.
Tertiary amines do not show any of these signals since they do not have a hydrogen bound to nitrogen.
Amines similar to OH
N-H broadens with Hydrogen bonding

36. IR: Amines Examples

37. IR: Carbonyls: C=O Aldehydes
Carbonyls in general:
Strong, sharp C=O peak 1670 to 1780 cm1
Conjugation lowers stretching frequency

38. IR: C=O: Aldehydes

39. IR: C=O: Ketones
Conjugation with a double bond or benzene ring lowers the stretching frequency by 30 to 40 cm-1.

40. IR: C=O: Ketones Ring strain increases frequency:
Incorporation of the carbonyl group in a small ring (5, 4 or 3-membered), raises the stretching frequency.

41. IR: C=O: Esters 1735 cm1 in saturated esters
Electron donating O increased the frequency
1715 cm1 in esters next to aromatic ring or a double bond
Conjugation decreases the frequency

42. IR: Carboxylic Acids

43. Learning Check:
32. Which of the following represents cyclohexane and which cyclohexene?

44. Solution:
32. Which of the following represents cyclohexane and which cyclohexene?
=C- H
-C-H
-C=C-

45. Learning Check:
41.Propose a structure for the following unknown hydrocarbon:

46. Solution:
68-53 =15 loss of CH3
41.Propose a structure for the following unknown hydrocarbon:
M-1=67
C5H8
2 deg of unsat
M=68
≡C- H
-C≡C-
-C-H

47. Learning Check:
42. Propose a structure for the following unknown hydrocarbon:

48. Solution:
42. Propose a structure for the following unknown hydrocarbon:
70-55 =15 loss of CH3
C5H10
M=70
-C-H
=C-H
-C=C-

49. base peak parent peak fragment analyte none of these
In mass spectrometry, what term is used to describe the ion that results from the ejection of one electron from a molecule?
base peak
parent peak
fragment
analyte
none of these

50. What quantity is detected by mass spectrometry?
the energy of a molecule
the number of electrons ejected from a molecule
the number of ions of a particular mass to charge ratio
the number of electrons needed to ionize a molecule
the number of hydrogen atoms in a molecule

51. High-resolution mass spectrometry enables one to determine the molecular formula of a molecule.
True
False

52. High-resolution mass spectrometry would allow one to distinguish between the following molecules.
True
False

53. Which of the given C6H12O isomers would be expected to produce an m/z peak at M+-18?3 4 2. 1. 4. 3.

54. Which of the following reasons explains why some mass spectrum peaks are larger than others?
The larger peaks represent fragments that are more stable.
The larger peaks represent fragments that are less stable.
Cations tend to give larger peaks than radical cations.
Radical cations tend to give larger peaks than cations.
None of these

55. Which of the following compounds is most likely to undergo a McLafferty rearrangement?1. 2. 1 2 3 4
Picture Choice 5 3. 4. 5.

56. Frequency is commonly reported in units of:
Joules nm
amu Hz
m/z

57. The higher the wavenumber of a molecular vibration, the lower the energy of the infrared radiation needed to stimulate it.
True
False

58. Which type of electromagnetic radiation possesses the highest energy?
IR light
UV light
microwaves
visible light
FM radio waves

59. Infrared spectroscopy is based on _____ excitation.
electronic
rotational
nuclear
vibrational

60. aldehyde and alkene alkene and alcohol alcohol and nitrile
What functional groups are most likely present in a compound whose IR spectrum shows absorbances at 2217 and 1648 cm-1?
aldehyde and alkene
alkene and alcohol
alcohol and nitrile
alkyne and ketone
nitrile and alkene

61. (R)-2-pentanol and (S)-2-pentanol give identical IR spectra.
True
False

62. Which of the following compounds will have its carbonyl absorb at the lowest frequency in IR spectroscopy? 1. 2. 1 2 3 4 5 3. 4. 5.

63. What approximate frequency range is considered the fingerprint region in infrared spectroscopy?
4000 – 3500 cm-1
3500 – 3000 cm-1
3000 – 2000 cm-1
2000 – 1500 cm-1
1500 – 500 cm-1

64. (E)-2-Butene and (Z)-2-butene give identical IR spectra.
True
False